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If a car traveled from Townsend to Smallville at an average

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If a car traveled from Townsend to Smallville at an average [#permalink] New post 07 Nov 2011, 13:46
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If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The distance from Townsend to Smallville is 165 miles.

How can the answer be A folks?
[Reveal] Spoiler: OA

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Re: Townsend to Smallville [#permalink] New post 07 Nov 2011, 15:03
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enigma123 wrote:
If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The distance from Townsend to Smallville is 165 miles.

How can the answer be A folks?



Lets distance be x miles between 2 cities (lets say T and S).
From T to S speed = 40 mph so time taken = x/40 hrs
From S to T spees = 60 mph (50% less time) so time taken = x/60 hrs

Total time taken = x/40 + x/60 = x/24 hrs
Total distance travelled = 2x miles

So avg speed = 2x/(x/24) = 48 mph

Instead of x you can solve this by taking 40 miles as distance.

So statement 1 is sufficient.
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Re: Townsend to Smallville [#permalink] New post 08 Nov 2011, 09:30
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enigma123 wrote:
If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The distance from Townsend to Smallville is 165 miles.

How can the answer be A folks?


A couple of basics:
1. Average Speed = Total Distance/ Total Time
2. Average Speed = Weighted Average of different speeds where 'time taken' is the weight
3. If equal distances (say d) are traveled at different speeds, a and b, average speed = 2d/[d/a + d/b] = 2ab/(a+b)
In this case, average speed is independent of 'd', the distance.
4. If speeds a and b are maintained for equal time intervals, say t,
average speed = (at+bt)/2t = (a+b)/2
In this case, average speed is independent of 't', the time interval.

Given: Car travels equal distances. Speed in first case is 40 mph. We need the speed in the second case to get the average speed (given by 2*40*b/(40+b))

(1) Ratio of time in the two cases
T to S: S to T= 3:2
Then ratio of speed in the two cases = 2:3
We know the speed from T to S = 40
Then speed from S to T must be 60
Since we have the value of b (= 60) i.e. the speed while coming back, we can easily find the average speed..
Sufficient.

(2) As we discussed, distance traveled doesn't affect average speed in this case. Not sufficient

Answer (A)
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If a car traveled from Townsend to Smallville at an average [#permalink] New post 11 Jul 2013, 07:38
If a car travelled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend along the same route later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The route between Townsend and Smallville is 165 miles long.

[Reveal] Spoiler:
Can someone provide a detailed analysis for statement A. Thanks

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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 11 Jul 2013, 07:48
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fozzzy wrote:
[Reveal] Spoiler:
Can someone provide a detailed analysis for statement A. Thanks


hi,

let say distance between T and S be X miles
time take to cover from T-S=X/40

statement 1==>time for T-S=1.5(time from S-T)
THEREFORE==>X/40=1.5(T)
THEREFORE ==>T=X/60

Average speed =total dist/total time

==>total dist=X+X=2X
AVG SPEED=2X/(X/40+X/60)=48
HENCE SUFFICIENT.

HOPE IT HELPS
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Last edited by blueseas on 11 Jul 2013, 07:58, edited 1 time in total.
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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 11 Jul 2013, 07:55
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If a car travelled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend along the same route later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.

Time from T to S 3/2 of that from S to T, thus
Speed from T to S 2/3 of that from S to T. Thus speed from S to T = 60 mph.

Now, (average \ speed)=\frac{(total \ distance)}{(total \ time)}=\frac{2d}{\frac{d}{40}+\frac{d}{60}}, where d is the distance between S and T --> d reduces --> we can get the average time. Sufficient.

(2) The route between Townsend and Smallville is 165 miles long. We know nothing about the speed from S to T. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 11 Jul 2013, 07:56
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shaileshmishra wrote:
fozzzy wrote:
[Reveal] Spoiler:
Can someone provide a detailed analysis for statement A. Thanks


hi,

let say distance between T and S be X miles
time take to cover from T-S=X/40

statement 1==>time for T-S=1.5(time from S-T)
THEREFORE==>X/40=1.5(T)
THEREFORE ==>T=X/60

Average speed =total dist/total time

==>total dist=X+X=2X
AVG SPEED=2X/(X/40+X/60)=50
HENCE SUFFICIENT.

HOPE IT HELPS


Small correction: the average speed would be 48 mph, not 50.
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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 11 Jul 2013, 07:59
Bunuel wrote:
[]

Small correction: the average speed would be 48 mph, not 50.


thanks bunuel :)
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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 29 Jul 2013, 11:26
If a car travelled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend along the same route later that evening, what was the average speed for the entire trip?

Distance/Time = average speed

D/T = 40

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.

We're looking for the total time here. In other words, we want to know the total distance/total time (the distance to and from and the speed to and from) We know that the distance is the same for both trips so that can be represented as 2d. Because we don't know the distance, to find time we will have to get distance/speed = time.

Total distance = 2d
Total time = time from T to S = time from S to T. Time = distance/speed. We know that the speed from T to S was 50% more than from S to T which means that the speed from S to T was 50% greater (i.e. 40 + 40*.5 = 60MPH)

Total Distance/Total Time = total average
2d/([d/40] + [d/60]) = total average
2d/([3d/120 + [2d/120]) = total average
2d/(5d/120) = total average
2d * 120/5d
240d/5d
d=48

If we plug 48 into d/t = 40 then we can get a value for t.
SUFFICIENT

(2) The route between Townsend and Smallville is 165 miles long.
We know that the distance is 165 which means that the round trip is 330 miles. The problem is, we are looking for average speed and we have no idea what speed was done on the return trip.
INSUFFICIENT

(B)
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Re: If a car traveled from Townsend to Smallville at an average [#permalink] New post 19 Nov 2013, 21:29
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Re: If a car traveled from Townsend to Smallville at an average   [#permalink] 19 Nov 2013, 21:29
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